f(w) f(a) = 1 2πi w a Proof. There exists a number r such that the disc D(a,r) is contained in I(γ). For any ǫ < r, w a dw

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1 Proof[section] 5. Cauchy integral formula Theorem 5.. Suppose f is holomorphic inside and on a positively oriented curve. Then if a is a point inside, f(a) = w a dw. Proof. There exists a number r such that the disc D(a,r) is contained in I(). For any ǫ < r, w a dw = (a;ǫ) by the Deformation Theorem. So w a dw f(a) = w a dw (a;ǫ) 2π f(a+ǫe iθ ) f(a) = iǫe iθ dθ 0 ǫe iθ (2π) sup f(a+ǫe iθ ) f(a) 2π θ [0,2π] f(a) dw w a The right hand side tends to 0 as ǫ 0. So the left hand side is Liouville s Theorem. Theorem 5.2. If f is holomorphic on C and is bounded(in other words there exists M for which f(z) < M for all z) then f is constant. Proof. Suppose M for all w C. Fix a and b in C. Take R 2max{ a, b } so that w a R/2 and w b R/2 when w = R. By Cauchy s integral formula with = (0;R), So f(a) f(b) = a b 2π ( w a w b (w a)(w b) dw. f(a) f(b) a b 2πRM 2π (R/2) 2 bytheestimation Theorem. SinceRisarbitrarily large, LHS = 0. ) dw

2 Fundamental theorem of algebra. Theorem 5.3. Let p be a non-constant polynomial with constant coefficients. Then there exists w C such that p(w) = 0. Proof. Suppose not. Then p(z) 0 for all z. Since p(z) as z, there exists R such that / p(z) < for z > R. On D(0, R), /p(z) is continuous and hence bounded. Hence /p(z) is bounded on C. It is also holomorphic, so constant by Liouville Cauchy s formula. Theorem 5.4. Suppose f is holomorphic inside and on a positively oriented contour. Let a lie inside. Then f (n) (a) exists for n =,2,... and f (n) (a) = n! (w a) n+dw. Corollary 5.5. If f is holomorphic in an open set G, then f has derivatives of all orders in G. Proof. For n = 0 this is Cauchy s integral formula. We assume it is true for n = k and prove it for n = k +. By deformation theorem, we may replace by (a;2r). Take h < r. By Cauchy s formula for n = k, ( dw f (k) (a+h) f (k) (a) = k! ( (k +)! = (w a h) k+ (w ζ) k 2 dζ [a,a+h] ) dw (by Fundamental Theorem of Calculus for f(z) = ). z k+ Define F(h) = f(k) (a+h) f (k) (a) (k +)! = h = (k +2)! h ( (w a) k+ (w a) (k+2)dw ζ [a,a+h]( (w v) k 3 dv ) dz v [a,ζ] ) dw Since f is holomorphic (and hence continuous), it is bounded by some M on (since is compact). For v [a,ζ], ζ [a,h], w v r for all w. Also zeta a h. Hence by the estimation theorem So F(h) 0 as h 0. F(h) (k +2)! 2π h M h 2 4πr. rk+3 )

3 Theorem 5.6 (Morera s Theorem). Suppose f is continuous in an open set G and f(z)dz = 0 for all triangles in G. Then f is holomorphic on G. Proof. Let a G. Choose r so that D(a;r) G. Since D(a;r) is convex, the Antiderivative Theorem implies that there exists a holomorphic function F such that F = f. Since F is holomorphic on D(a,r), so is f. Since a is arbitrary, f is holomorphic on G. Example 5.. Use of Cauchy integral formula:. z 2 z 2 + dz (i;) We need to write the integrand as f(z) where f is holomorphic at a. z a Take f(z) = z2 z +i since z 2 + = (z +i)(z i). Then (i;) f(z) dz = f(a) z a so z 2 (z +i)(z i) dz = z 2 z +i = π 2. e z dz (0;) z 3 Rewrite as f(z) dz which equals 2πf (n) (a). We check that f is (z a) n+ holomorphic inside and on a contour enclosing a. Take f(z) = e z,n = 2 f (z) = f (z) = e z so the integral is 2! f(2) (0) = πi 3. Re(z) (0;) z /2 dz cannot be evaluated directly using the Cauchy integral formula since Re(z) is not a holomorphic function of z. But (0;) is the unit circle so Re(z) = (z + z)/2 and if z =, z = z so Re(z) = (z +z )/2. So on (0;) Re(z) z /2 = z 2 + 2z(z /2) z = 2(z /2) + z(2z ) = 2 (z /2)+/2 z /2 + A z + B 2z 3

4 4 = z = z Solving, A = and B = 2A = 2. So our expression is So by Cauchy s formula + A(2z )+Bz z(2z ) + (2A+B)z A z(2z ) = z + ( 2 + z (2z ) = 2 z + 5 4(z /2) (0;) 5.4. Poisson integral formula. Re(z) dz = /4 z /2 Theorem 5.7. Suppose f is holomorphic inside and on (0;). Then f(re iθ = 2π r 2 )dt. 2π 0 2rcos(θ t)+r 2f(eit Proof. Fix z = re iθ. Apply Cauchy integral formula to g(z) = f(z)φ(z) where Note that φ(z) =. Then = w2 φ(w) = r2 w z. f(z) = f(z)φ(z) (0;) (w z)( w z) dw as required. = r2 2π 0 f(e it )ie it dt (e it re iθ )( re it e iθ )

5 5.5. Power Series. Theorem 5.8. (Taylor s Theorem) Suppose f is holomorphic on D(a; R). Then there exist constants c n so that f(z) = c n (z a) n n=0 and c n = (w a) n+dw = f(n) (a) n! where is the circle (a;r) for 0 < r < R. Proof. Fix z D(a;R) and choose r so that z a < r < R Take = (a;r). Then f(z) = dw (by Cauchy integral formula). w z Since z a < w a for all w, w z = Now apply the binomial expansion: f(z) = w a n=0 z a w a (z a) n (w a) n+dw Since is compact and f is continuous, f is bounded. So for some constant M, (z a) n (w a) n+ M r n (z a) := M n. r Since z a < r, n M n converges. So (by uniform convergence theorem) we may interchange summation and integration. So f(z) = ( ) (w a) n+dw (z a) n. n=0 Hence the theorem follows by Cauchy s formula for derivatives. Example 5.2. Let f be holomorphic on C. Prove that if there are M > 0 and K > 0, 0 < k Z such that f(z) M z k for z K, then f is a polynomial of degree k. Proof. By Taylor s theorem f has a power series expansion f(z) = n=0 c nz n in any disk with center 0. c n = f(z)z n dz. (0;R) 5

6 6 So if R K, c n 2π sup{ f(z)z n : z = R} times the length of (0;R). So this is 2π MRk n (2πR). Since R can be chosen arbitrarily large, we must have c n = 0 for n > k. Thus f is a polynomial of degree k. Proposition 5.. Let f(z) = n=0 a nz n and g(z) = n=0 b nz n. Then f(z)g(z) = c n z n where c n = n r=0 a rb n r If the radius of convergence of the power series for f is R and that for g is R 2 then the radius of convergence of the power series for fg is at least the minimum of R and R 2. Example 5.3. The power series for exp(z) has infinite radius of convergence (by the Ratio Test). n= Zeros of holomorphic functions. Definition 5.9. a is an isolated zero of f if there is ǫ > 0 such that D (a,ǫ) contains no zeros of f. Theorem 5.0. (Identity Theorem) Suppose G is a region and f is holomorphic on G. If the set of zeros of f on G has a limit point in G then f is zero everywhere in G. (Equivalently the zeros of f are isolated unless f is zero everywhere.) Proof. Leta Gwithf(a) = 0. ByTaylor stheorem(z) = n=0 c n(z a) n for z a r. Either all c n = 0 (which implies f = 0 everywhere in D(a;r)) or there is a smallest m > 0 with c m 0. The series n=0 c n+m(z a) n has radius of convergence at least r, and defines a function g which is continuous in D(a;r) because a power series defines a holomorphic function inside its radius of convergence. Since g(a) 0 and g is continuous at a, g(z) 0 for z in some disk D(a;ǫ). In the punctured disk D(a,ǫ\{a}), f(z) = (z a) m g(z) is never zero. So a is not a limit point of the set of zeros of f. Hence if f(a) = 0, either f = 0 in some disc, or a is not a limit point (in other words there is some disc where f(z) 0 except for z = a). Sketch proof that if there is a limit point of the set of zeros in G, then f = 0: everywhere in G: Show that the set E of limit points is

7 contained in the set Z(f) of the set of zeros and both E and G\E are open (hence either E = G or E is the empty set since G is connected). (i) Suppose there is a limit point a for which f(a) 0. Since it is a limit point, there are disks of radius /n containing points a n with f(a n ) = 0 and a a n < /n. Since f is continuous this implies f(a) = 0 contrary to hypothesis. So (ii) To show both E and G \ E are open: Suppose a E. Then f = 0 in some disk D around a (by part ). So this disk D E. So E is open. To show G\E is open: Take a G\E. Since a is not a limit point of Z(f) there is a disc D(a;r) in which f is never zero. By (i) E Z(f). Hence D(a;r) G\E Maximum Modulus Theorem. Theorem 5.. Suppose f is holomorphic on D(a,R) with f(z) f(a) for all z D(a,R). Then f is constant. Theorem 5.2. Suppose G is a bounded region, f is holomorphic in G and continuous on Ḡ. Then f attains its maximum on the boundary of G, in other words on G = Ḡ\G. Proof of Theorem 5.: Choose 0 < r < R. By the Cauchy integral formula f(a) = f(z) z a dz ((t) = re it ) Hence = f(a) 2i (by hypothesis on f). 2π 0 2π 0 = 2π 2π 0 (a;r ( f(a+reit ) (re it )dt rie it f(a+re it )dt. f(a+re it ) dt f(a). [ f(a) f(a+re it ) ] dt = 0. (The integrand is continuous and 0, so it must be equal to 0.) This is true for all r < R. So f is constant in D(a;R). So f is also constant in D(a;R). Proof of Theorem 5.2: Ḡ is closed and bounded, so on Ḡ, f is bounded (as it is continuous) andattainsitsmaximumvaluem atsomepointonḡ. Assume f does 7

8 8 not attain M on the boundary G. Then f(a) = M for some a G. By part (a), f is constant on some disk D(a;R) G. Hence f is constant in G, by the identity theorem. By continuity, f is constant on Ḡ, so it attains its maximum on G, contradicting the hypothesis. Example 5.4 (Examples). () Is it possible to have a holomorphic function that is equal to 0 everywhere on the real axis? Answer: No, since the Identity Theorem says that if the set of zeros has a limit point then f is zero everywhere. (A limit point is a point p for which any disc containing p, no matter how small, will contain some points z where f(z) = 0.) (2) Is it possible to have a holomorphic function which is equal to when z = and equal to when z =? 2n 2n+ Answer: No, since the set of points { } has a limit point 2n (namely 0) and f = on those points. Likewise the set of points { } has a limit point (namely 0) and f = on 2n+ those points. Looking at the points { } we conclude f = 2n everywhere (by the identity theorem). Likewise looking at the points { } we conclude f = everywhere (by the identity 2n+ theorem). This is a contradiction.